ch01

Prévia do material em texto

PROBLEM 1.1 
 
KNOWN: Heat rate, q, through one-dimensional wall of area A, thickness L, thermal 
conductivity k and inner temperature, T1. 
 
FIND: The outer temperature of the wall, T2. 
 
SCHEMATIC: 
 
 
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, 
(3) Constant properties. 
 
ANALYSIS: The rate equation for conduction through the wall is given by Fourier’s law, 
 
q q q A = -k dT
dx
A = kA T T
Lcond x x
1 2
= = ′′ ⋅ ⋅
−
. 
 
Solving for T2 gives 
 
 T T
q L
kA2 1
cond
= − . 
 
Substituting numerical values, find 
 
 T C - 3000W 0.025m
0.2W / m K 10m2 2
=
×
⋅ ×
415$ 
 
 T C - 37.5 C2 = 415
$ $
 
 
 T C.2 = 378
$
 < 
 
COMMENTS: Note direction of heat flow and fact that T2 must be less than T1. 
PROBLEM 1.2 
 
KNOWN: Inner surface temperature and thermal conductivity of a concrete wall. 
 
FIND: Heat loss by conduction through the wall as a function of ambient air temperatures ranging from 
-15 to 38°C. 
 
SCHEMATIC: 
 
 
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state conditions, (3) 
Constant properties, (4) Outside wall temperature is that of the ambient air. 
 
ANALYSIS: From Fourier’s law, it is evident that the gradient, xdT dx q k′′= − , is a constant, and 
hence the temperature distribution is linear, if xq′′ and k are each constant. The heat flux must be 
constant under one-dimensional, steady-state conditions; and k is approximately constant if it depends 
only weakly on temperature. The heat flux and heat rate when the outside wall temperature is T2 = -15°C 
are 
 
( ) 21 2
x
25 C 15 CdT T Tq k k 1W m K 133.3W m
dx L 0.30m
− −
−
′′ = − = = ⋅ =
$ $
. (1) 
 
2 2
x xq q A 133.3W m 20m 2667 W′′= × = × = . (2) < 
 
Combining Eqs. (1) and (2), the heat rate qx can be determined for the range of ambient temperature, -15 
≤ T2 ≤ 38°C, with different wall thermal conductivities, k. 
-20 -10 0 10 20 30 40
Ambient air temperature, T2 (C)
-1500
-500
500
1500
2500
3500
H
ea
t l
os
s,
 q
x 
(W
)
Wall thermal conductivity, k = 1.25 W/m.K
k = 1 W/m.K, concrete wall
k = 0.75 W/m.K
 
 
For the concrete wall, k = 1 W/m⋅K, the heat loss varies linearily from +2667 W to -867 W and is zero 
when the inside and ambient temperatures are the same. The magnitude of the heat rate increases with 
increasing thermal conductivity. 
 
COMMENTS: Without steady-state conditions and constant k, the temperature distribution in a plane 
wall would not be linear. 
PROBLEM 1.3 
 
KNOWN: Dimensions, thermal conductivity and surface temperatures of a concrete slab. Efficiency 
of gas furnace and cost of natural gas. 
 
FIND: Daily cost of heat loss. 
 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) Steady state, (2) One-dimensional conduction, (3) Constant properties. 
 
ANALYSIS: The rate of heat loss by conduction through the slab is 
 
 ( ) ( )1 2T T 7 Cq k LW 1.4 W / m K 11m 8m 4312 W
t 0.20 m
− °
= = ⋅ × = < 
 
The daily cost of natural gas that must be combusted to compensate for the heat loss is 
 
 ( ) ( )gd 6f
q C 4312 W $0.01/ MJC t 24h / d 3600s / h $4.14 / d
0.9 10 J / MJη
×
= ∆ = × =
×
 
<
 
 
COMMENTS: The loss could be reduced by installing a floor covering with a layer of insulation 
between it and the concrete. 
 
PROBLEM 1.4
KNOWN: Heat flux and surface temperatures associated with a wood slab of prescribed
thickness.
FIND: Thermal conductivity, k, of the wood.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state
conditions, (3) Constant properties.
ANALYSIS: Subject to the foregoing assumptions, the thermal conductivity may be
determined from Fourier’s law, Eq. 1.2. Rearranging,
( )
L W 0.05mk=q 40 
T T m 40-20 C
x 21 2
′′ =
−
�
k = 0.10 W / m K.⋅ <
COMMENTS: Note that the °C or K temperature units may be used interchangeably when
evaluating a temperature difference.
PROBLEM 1.5
KNOWN: Inner and outer surface temperatures of a glass window of prescribed dimensions.
FIND: Heat loss through window.
SCHEMATIC:
ASSUMPTIONS: (1) One-dimensional conduction in the x-direction, (2) Steady-state
conditions, (3) Constant properties.
ANALYSIS: Subject to the foregoing conditions the heat flux may be computed from
Fourier’s law, Eq. 1.2.
( )
T Tq k 
L
15-5 CWq 1.4 
m K 0.005m
q 2800 W/m .
1 2
x
x
2
x
−
′′ =
′′ =
⋅
′′ =
�
Since the heat flux is uniform over the surface, the heat loss (rate) is
q = qx A
q = 2800 W / m2 3m2
′′ ×
×
q = 8400 W. <
COMMENTS: A linear temperature distribution exists in the glass for the prescribed
conditions.
PROBLEM 1.6 
 
KNOWN: Width, height, thickness and thermal conductivity of a single pane window and 
the air space of a double pane window. Representative winter surface temperatures of single 
pane and air space. 
 
FIND: Heat loss through single and double pane windows. 
 
SCHEMATIC: 
 
 
 
 
ASSUMPTIONS: (1) One-dimensional conduction through glass or air, (2) Steady-state 
conditions, (3) Enclosed air of double pane window is stagnant (negligible buoyancy induced 
motion). 
 
ANALYSIS: From Fourier’s law, the heat losses are 
 
Single Pane: ( )T T 35 C21 2q k A 1.4 W/m K 2m 19,600 Wg g L 0.005m−= = ⋅ =$ 
 
Double Pane: ( )T T 25 C21 2q k A 0.024 2m 120 Wa a L 0.010 m−= = =$ 
 
COMMENTS: Losses associated with a single pane are unacceptable and would remain 
excessive, even if the thickness of the glass were doubled to match that of the air space. The 
principal advantage of the double pane construction resides with the low thermal conductivity 
of air (~ 60 times smaller than that of glass). For a fixed ambient outside air temperature, use 
of the double pane construction would also increase the surface temperature of the glass 
exposed to the room (inside) air. 
 
PROBLEM 1.7
KNOWN: Dimensions of freezer compartment. Inner and outer surface temperatures.
FIND: Thickness of styrofoam insulation needed to maintain heat load below prescribed
value.
SCHEMATIC:
ASSUMPTIONS: (1) Perfectly insulated bottom, (2) One-dimensional conduction through 5
walls of area A = 4m2, (3) Steady-state conditions, (4) Constant properties.
ANALYSIS: Using Fourier’s law, Eq. 1.2, the heat rate is
q = q A = k T
L
 Atotal′′ ⋅
∆
Solving for L and recognizing that Atotal = 5×W
2
, find
L = 5 k T W
q
2∆
( ) ( )5 0.03 W/m K 35 - -10 C 4m
L = 
500 W
2 × ⋅  �
L = 0.054m = 54mm. <
COMMENTS: The corners will cause local departures from one-dimensional conduction
and a slightly larger heat loss.
PROBLEM 1.8 
 
KNOWN: Dimensions and thermal conductivity of food/beverage container. Inner and outer 
surface temperatures. 
 
FIND: Heat flux through container wall and total heat load. 
 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer through bottom 
wall, (3) Uniform surface temperatures and one-dimensional conduction through remaining 
walls. 
 
ANALYSIS: From Fourier’s law, Eq. 1.2, the heat flux is 
 
 
( )0.023 W/m K 20 2 CT T 22 1q k 16.6 W/m
L 0.025 m
⋅ −
−
′′ = = =
$
 < 
 
Since the flux is uniform over each of the five walls through which heat is transferred, the 
heat load is 
 
 ( )q q A q H 2W 2W W Wtotal 1 2 1 2′′ ′′  = × = + + ×  
 
 ( ) ( )2q 16.6 W/m 0.6m 1.6m 1.2m 0.8m 0.6m 35.9 W = + + × =  < 
 
COMMENTS: The corners and edges of the container create local departures from one-
dimensional conduction, which increasethe heat load. However, for H, W1, W2 >> L, the 
effect is negligible. 
 
PROBLEM 1.9 
 
KNOWN: Masonry wall of known thermal conductivity has a heat rate which is 80% of that 
through a composite wall of prescribed thermal conductivity and thickness. 
 
FIND: Thickness of masonry wall. 
 
SCHEMATIC: 
 
 
ASSUMPTIONS: (1) Both walls subjected to same surface temperatures, (2) One-
dimensional conduction, (3) Steady-state conditions, (4) Constant properties. 
 
ANALYSIS: For steady-state conditions, the conduction heat flux through a one-dimensional 
wall follows from Fourier’s law, Eq. 1.2, 
 
 ′′q = k T
L
∆
 
 
where ∆T represents the difference in surface temperatures. Since ∆T is the same for both 
walls, it follows that 
 
 L = L 
k
k
 
q
q1 2
1
2
2
1
⋅
′′
′′
. 
 
With the heat fluxes related as 
 
 ′′ = ′′q 0.8 q1 2 
 
 L = 100mm 0.75 W / m K
0.25 W / m K
 
1
0.8
 = 375mm.1
⋅
⋅
× < 
 
COMMENTS: Not knowing the temperature difference across the walls, we cannot find the 
value of the heat rate. 
 
PROBLEM 1.10 
 
KNOWN: Thickness, diameter and inner surface temperature of bottom of pan used to boil 
water. Rate of heat transfer to the pan. 
 
FIND: Outer surface temperature of pan for an aluminum and a copper bottom. 
 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) One-dimensional, steady-state conduction through bottom of pan. 
 
ANALYSIS: From Fourier’s law, the rate of heat transfer by conduction through the bottom 
of the pan is 
 
 
T T1 2q kA
L
−
= 
 
Hence, 
 
 
qLT T1 2 kA
= + 
 
where ( )22 2A D / 4 0.2m / 4 0.0314 m .pi pi= = = 
 
Aluminum: ( )( )
600W 0.005 m
T 110 C 110.40 C1 2240 W/m K 0.0314 m
= + =
⋅
$ $
 
 
Copper: ( )( )
600W 0.005 m
T 110 C 110.25 C1 2390 W/m K 0.0314 m
= + =
⋅
$ $
 
 
COMMENTS: Although the temperature drop across the bottom is slightly larger for 
aluminum (due to its smaller thermal conductivity), it is sufficiently small to be negligible for 
both materials. To a good approximation, the bottom may be considered isothermal at T ≈ 
110 °C, which is a desirable feature of pots and pans. 
 
PROBLEM 1.11
KNOWN: Dimensions and thermal conductivity of a chip. Power dissipated on one surface.
FIND: Temperature drop across the chip.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Constant properties, (3) Uniform heat
dissipation, (4) Negligible heat loss from back and sides, (5) One-dimensional conduction in
chip.
ANALYSIS: All of the electrical power dissipated at the back surface of the chip is
transferred by conduction through the chip. Hence, from Fourier’s law,
P = q = kA T
t
∆
or
( )
t P 0.001 m 4 WT = 
kW 150 W/m K 0.005 m2 2
⋅ ×∆ =
⋅
∆T = 1.1 C.$ <
COMMENTS: For fixed P, the temperature drop across the chip decreases with increasing k
and W, as well as with decreasing t.
PROBLEM 1.12 
 
KNOWN: Heat flux gage with thin-film thermocouples on upper and lower surfaces; output 
voltage, calibration constant, thickness and thermal conductivity of gage. 
 
FIND: (a) Heat flux, (b) Precaution when sandwiching gage between two materials. 
 
SCHEMATIC: 
 
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional heat conduction in gage, 
(3) Constant properties. 
 
ANALYSIS: (a) Fourier’s law applied to the gage can be written as 
 
 ′′q = k T
x
∆
∆
 
 
and the gradient can be expressed as 
 
 
∆
∆
∆T
x
 = 
E / N
SABt
 
 
where N is the number of differentially connected thermocouple junctions, SAB is the Seebeck 
coefficient for type K thermocouples (A-chromel and B-alumel), and ∆x = t is the gage 
thickness. Hence, 
 
 ′′q = k E
NSABt
∆
 
 
 ′′
⋅ × ×
× × × ×
q = 1.4 W / m K 350 10
-6
 V
5 40 10-6 V / C 0.25 10-3 m
 = 9800 W / m2
$
. < 
 
(b) The major precaution to be taken with this type of gage is to match its thermal 
conductivity with that of the material on which it is installed. If the gage is bonded 
between laminates (see sketch above) and its thermal conductivity is significantly different 
from that of the laminates, one dimensional heat flow will be disturbed and the gage will 
read incorrectly. 
 
COMMENTS: If the thermal conductivity of the gage is lower than that of the laminates, 
will it indicate heat fluxes that are systematically high or low? 
 
PROBLEM 1.13 
 
KNOWN: Hand experiencing convection heat transfer with moving air and water. 
 
FIND: Determine which condition feels colder. Contrast these results with a heat loss of 30 W/m2 under 
normal room conditions. 
 
SCHEMATIC: 
 
 
ASSUMPTIONS: (1) Temperature is uniform over the hand’s surface, (2) Convection coefficient is 
uniform over the hand, and (3) Negligible radiation exchange between hand and surroundings in the case 
of air flow. 
 
ANALYSIS: The hand will feel colder for the condition which results in the larger heat loss. The heat 
loss can be determined from Newton’s law of cooling, Eq. 1.3a, written as 
 
 ( )sq h T T∞′′ = − 
 
For the air stream: 
 
 ( )2 2airq 40 W m K 30 5 K 1,400 W m′′  = ⋅ − − =  < 
 
For the water stream: 
 
 ( )2 2waterq 900 W m K 30 10 K 18,000 W m′′ = ⋅ − = < 
 
COMMENTS: The heat loss for the hand in the water stream is an order of magnitude larger than when 
in the air stream for the given temperature and convection coefficient conditions. In contrast, the heat 
loss in a normal room environment is only 30 W/m2 which is a factor of 400 times less than the loss in 
the air stream. In the room environment, the hand would feel comfortable; in the air and water streams, 
as you probably know from experience, the hand would feel uncomfortably cold since the heat loss is 
excessively high. 
PROBLEM 1.14 
 
KNOWN: Power required to maintain the surface temperature of a long, 25-mm diameter cylinder 
with an imbedded electrical heater for different air velocities. 
 
FIND: (a) Determine the convection coefficient for each of the air velocity conditions and display 
the results graphically, and (b) Assuming that the convection coefficient depends upon air velocity as 
h = CVn, determine the parameters C and n. 
 
SCHEMATIC: 
 
 
V(m/s) 1 2 4 8 12 
′Pe (W/m) 450 658 983 1507 1963 
h (W/m2⋅K) 22.0 32.2 48.1 73.8 96.1 
 
ASSUMPTIONS: (1) Temperature is uniform over the cylinder surface, (2) Negligible radiation 
exchange between the cylinder surface and the surroundings, (3) Steady-state conditions. 
 
ANALYSIS: (a) From an overall energy balance on the cylinder, the power dissipated by the 
electrical heater is transferred by convection to the air stream. Using Newtons law of cooling on a per 
unit length basis, 
 
 ( )( )e sP h D T Tpi ∞′ = − 
 
where eP′ is the electrical power dissipated per unit length of the cylinder. For the V = 1 m/s 
condition, using the data from the table above, find 
 ( ) 2h 450 W m 0.025m 300 40 C 22.0 W m Kpi= × − = ⋅$ < 
 
Repeating the calculations, find the convection coefficients for the remaining conditions which are 
tabulated above and plotted below. Note that h is not linear with respect to the air velocity. 
 
(b) To determine the (C,n) parameters, we plotted h vs. V on log-log coordinates. Choosing C = 
22.12 W/m2⋅K(s/m)n, assuring a match at V = 1, we can readily find the exponent n from the slope of 
the h vs. V curve. From the trials with n = 0.8, 0.6 and 0.5, we recognize that n = 0.6 is a reasonable 
choice. Hence, C = 22.12 and n = 0.6. < 
 
0 2 4 6 8 10 12
Air velocity, V (m/s)
20
40
60
80
100
Co
ef
fic
ie
nt
, h
 (W
/m
^2
.K
)
Data,smooth curve, 5-points
 
1 2 4 6 8 10
Air velocity, V (m/s) 
10
20
40
60
80
100
Co
ef
fic
ie
nt
, h
 (W
/m
^2
.K
) 
Data , smooth curve, 5 points
h = C * V^n, C = 22.1, n = 0.5
n = 0.6
n = 0.8
 
 
PROBLEM 1.15
KNOWN: Long, 30mm-diameter cylinder with embedded electrical heater; power required
to maintain a specified surface temperature for water and air flows.
FIND: Convection coefficients for the water and air flow convection processes, hw and ha,
respectively.
SCHEMATIC:
ASSUMPTIONS: (1) Flow is cross-wise over cylinder which is very long in the direction
normal to flow.
ANALYSIS: The convection heat rate from the cylinder per unit length of the cylinder has
the form
( ) ( )q = h D T Tsπ′ − ∞
and solving for the heat transfer convection coefficient, find
( )
qh = .
D T Tsπ
′
− ∞
Substituting numerical values for the water and air situations:
Water
( )
28 10 W/mh = = 4,570 W/m K
 0.030m 90-25 C
3 2
w
π
×
⋅
×
�
<
Air
( )
400 W/mh = 65 W/m K.
 0.030m 90-25 C
2
a
π
= ⋅
×
�
<
COMMENTS: Note that the air velocity is 10 times that of the water flow, yet
hw ≈ 70 × ha.
These values for the convection coefficient are typical for forced convection heat transfer with
liquids and gases. See Table 1.1.
PROBLEM 1.16
KNOWN: Dimensions of a cartridge heater. Heater power. Convection coefficients in air
and water at a prescribed temperature.
FIND: Heater surface temperatures in water and air.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) All of the electrical power is transferred
to the fluid by convection, (3) Negligible heat transfer from ends.
ANALYSIS: With P = qconv, Newton’s law of cooling yields
( ) ( )P=hA T T h DL T T
PT T .
h DL
s s
s
π
π
− = −
= +
∞ ∞
∞
In water,
T C + 2000 W
5000 W / m K 0.02 m 0.200 ms 2
=
⋅ × × ×
20$
π
T C + 31.8 C = 51.8 C.s = 20$ $ $ <
In air,
T C + 2000 W
50 W / m K 0.02 m 0.200 ms 2
=
⋅ × × ×
20$
π
T C + 3183 C = 3203 C.s = 20
$ $ $ <
COMMENTS: (1) Air is much less effective than water as a heat transfer fluid. Hence, the
cartridge temperature is much higher in air, so high, in fact, that the cartridge would melt.
(2) In air, the high cartridge temperature would render radiation significant.
PROBLEM 1.17 
 
KNOWN: Length, diameter and calibration of a hot wire anemometer. Temperature of air 
stream. Current, voltage drop and surface temperature of wire for a particular application. 
 
FIND: Air velocity 
 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer from the wire by 
natural convection or radiation. 
 
ANALYSIS: If all of the electric energy is transferred by convection to the air, the following 
equality must be satisfied 
 
 ( )P EI hA T Telec s= = − ∞ 
 
where ( ) 5 2A DL 0.0005m 0.02m 3.14 10 m .pi pi −= = × = × 
 
Hence, 
 
 ( ) ( )
EI 5V 0.1A 2h 318 W/m K
5 2A T Ts 3.14 10 m 50 C
×
= = = ⋅
−
− ∞ × $
 
 
 ( )25 2 5 2V 6.25 10 h 6.25 10 318 W/m K 6.3 m/s− −= × = × ⋅ = < 
 
COMMENTS: The convection coefficient is sufficiently large to render buoyancy (natural 
convection) and radiation effects negligible. 
 
PROBLEM 1.18 
 
KNOWN: Chip width and maximum allowable temperature. Coolant conditions. 
 
FIND: Maximum allowable chip power for air and liquid coolants. 
 
SCHEMATIC: 
 
 
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer from sides and 
bottom, (3) Chip is at a uniform temperature (isothermal), (4) Negligible heat transfer by 
radiation in air. 
 
ANALYSIS: All of the electrical power dissipated in the chip is transferred by convection to 
the coolant. Hence, 
 
 P = q 
 
and from Newton’s law of cooling, 
 
 P = hA(T - T∞) = h W2(T - T∞). 
 
In air, 
 
 Pmax = 200 W/m
2
⋅K(0.005 m)2(85 - 15) ° C = 0.35 W. < 
 
In the dielectric liquid 
 
 Pmax = 3000 W/m
2
⋅K(0.005 m)2(85-15) ° C = 5.25 W. < 
 
COMMENTS: Relative to liquids, air is a poor heat transfer fluid. Hence, in air the chip can 
dissipate far less energy than in the dielectric liquid. 
PROBLEM 1.19 
 
KNOWN: Length, diameter and maximum allowable surface temperature of a power 
transistor. Temperature and convection coefficient for air cooling. 
 
FIND: Maximum allowable power dissipation. 
 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible heat transfer through base of 
transistor, (3) Negligible heat transfer by radiation from surface of transistor. 
 
ANALYSIS: Subject to the foregoing assumptions, the power dissipated by the transistor is 
equivalent to the rate at which heat is transferred by convection to the air. Hence, 
 
 ( )P q hA T Telec conv s= = − ∞ 
 
where ( ) ( )22 4 2A DL D / 4 0.012m 0.01m 0.012m / 4 4.90 10 m .pi pi − = + = × + = ×   
 
For a maximum allowable surface temperature of 85°C, the power is 
 
 ( ) ( )2 4 2P 100 W/m K 4.90 10 m 85 25 C 2.94 Welec −= ⋅ × − =$ < 
 
COMMENTS: (1) For the prescribed surface temperature and convection coefficient, 
radiation will be negligible relative to convection. However, conduction through the base 
could be significant, thereby permitting operation at a larger power. 
 
(2) The local convection coefficient varies over the surface, and hot spots could exist if there 
are locations at which the local value of h is substantially smaller than the prescribed average 
value. 
 
PROBLEM 1.20 
 
KNOWN: Air jet impingement is an effective means of cooling logic chips. 
 
FIND: Procedure for measuring convection coefficients associated with a 10 mm × 10 mm chip. 
 
SCHEMATIC: 
 
 
ASSUMPTIONS: Steady-state conditions. 
 
ANALYSIS: One approach would be to use the actual chip-substrate system, Case (a), to perform the 
measurements. In this case, the electric power dissipated in the chip would be transferred from the chip 
by radiation and conduction (to the substrate), as well as by convection to the jet. An energy balance for 
the chip yields elec conv cond radq q q q= + + . Hence, with ( )conv sq hA T T∞= − , where A = 100 
mm
2
 is the surface area of the chip, 
 
 ( )
elec cond rad
s
q q qh
A T T
∞
− −
=
−
 (1) 
 
While the electric power (qelec) and the jet (T∞ ) and surface (Ts ) temperatures may be measured, losses 
from the chip by conduction and radiation would have to be estimated. Unless the losses are negligible 
(an unlikely condition), the accuracy of the procedure could be compromised by uncertainties associated 
with determining the conduction and radiation losses. 
 
 A second approach, Case (b), could involve fabrication of a heater assembly for which the 
conduction and radiation losses are controlled and minimized. A 10 mm × 10 mm copper block (k ~ 400 
W/m⋅K) could be inserted in a poorly conducting substrate (k < 0.1 W/m⋅K) and a patch heater could be 
applied to the back of the block and insulated from below. If conduction to both the substrate and 
insulation could thereby be rendered negligible, heat would be transferred almost exclusively through the 
block. If radiation were rendered negligible by applying a low emissivity coating (ε < 0.1) to the surface 
of the copper block, virtually all of the heat would be transferred by convection to the jet. Hence, qcond 
and qrad may be neglected in equation (1), and the expression may be used to accurately determine h 
from the known (A) and measured (qelec, Ts , T∞ ) quantities. 
 
COMMENTS: Since convection coefficients associated with gas flows are generally small, concurrent 
heat transfer by radiation and/or conduction mustoften be considered. However, jet impingement is one 
of the more effective means of transferring heat by convection and convection coefficients well in excess 
of 100 W/m2⋅K may be achieved. 
PROBLEM 1.21
KNOWN: Upper temperature set point, Tset, of a bimetallic switch and convection heat
transfer coefficient between clothes dryer air and exposed surface of switch.
FIND: Electrical power for heater to maintain Tset when air temperature is T∞ = 50°C.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Electrical heater is perfectly insulated
from dryer wall, (3) Heater and switch are isothermal at Tset, (4) Negligible heat transfer from
sides of heater or switch, (5) Switch surface, As, loses heat only by convection.
ANALYSIS: Define a control volume around the bimetallic switch which experiences heat
input from the heater and convection heat transfer to the dryer air. That is,
( )
E - E = 0
q - hA T T 0.
outin
s setelec − =∞
� �
The electrical power required is,
( )q = hA T Ts setelec − ∞
( )q = 25 W/m K 30 10 m 70 50 K=15 mW.2 -6 2elec ⋅ × × − <
COMMENTS: (1) This type of controller can achieve variable operating air temperatures
with a single set-point, inexpensive, bimetallic-thermostatic switch by adjusting power levels
to the heater.
(2) Will the heater power requirement increase or decrease if the insulation pad is other than
perfect?
PROBLEM 1.22
KNOWN: Hot vertical plate suspended in cool, still air. Change in plate temperature with time at
the instant when the plate temperature is 225°C.
FIND: Convection heat transfer coefficient for this condition.
SCHEMATIC:
ASSUMPTIONS: (1) Plate is isothermal and of uniform temperature, (2) Negligible radiation
exchange with surroundings, (3) Negligible heat lost through suspension wires.
ANALYSIS: As shown in the cooling curve above, the plate temperature decreases with time. The
condition of interest is for time to. For a control surface about the plate, the conservation of energy
requirement is
( )
E - E = E
dT2hA T T Mc
dt
out stin
s s p− − =∞
� � �
where As is the surface area of one side of the plate. Solving for h, find
( )
Mc dTh=
2A T T dt
p
s s − ∞
( ) ( )
3.75 kg 2770 J/kg Kh= 0.022 K/s=6.4 W/m K
2 0.3 0.3 m 225 25 K
2
2
× ⋅
× ⋅
× × −
<
COMMENTS: (1) Assuming the plate is very highly polished with emissivity of 0.08, determine
whether radiation exchange with the surroundings at 25°C is negligible compared to convection.
(2) We will later consider the criterion for determining whether the isothermal plate assumption is
reasonable. If the thermal conductivity of the present plate were high (such as aluminum or copper),
the criterion would be satisfied.
PROBLEM 1.23 
 
KNOWN: Width, input power and efficiency of a transmission. Temperature and convection 
coefficient associated with air flow over the casing. 
 
FIND: Surface temperature of casing. 
 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) Steady state, (2) Uniform convection coefficient and surface temperature, (3) 
Negligible radiation. 
 
ANALYSIS: From Newton’s law of cooling, 
 
 ( ) ( )2s s sq hA T T 6 hW T T∞ ∞= − = − 
 
where the output power is η Pi and the heat rate is 
 
 ( )i o iq P P P 1 150hp 746 W / hp 0.07 7833Wη= − = − = × × = 
 
Hence, 
 
 
( )s 2 22
q 7833 WT T 30 C 102.5 C
6 hW 6 200 W / m K 0.3m
∞= + = ° + = °
× ⋅ ×
 
<
 
 
COMMENTS: There will, in fact, be considerable variability of the local convection coefficient 
over the transmission case and the prescribed value represents an average over the surface. 
 
PROBLEM 1.24 
 
KNOWN: Air and wall temperatures of a room. Surface temperature, convection coefficient 
and emissivity of a person in the room. 
 
FIND: Basis for difference in comfort level between summer and winter. 
 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) Person may be approximated as a small object in a large enclosure. 
 
ANALYSIS: Thermal comfort is linked to heat loss from the human body, and a chilled 
feeling is associated with excessive heat loss. Because the temperature of the room air is 
fixed, the different summer and winter comfort levels can not be attributed to convection heat 
transfer from the body. In both cases, the heat flux is 
 
Summer and Winter: ( ) 2 2q h T T 2 W/m K 12 C 24 W/mconv s′′ = − = ⋅ × =∞ $ 
 
However, the heat flux due to radiation will differ, with values of 
 
Summer: ( ) ( )4 4 8 2 4 4 4 4 2q T T 0.9 5.67 10 W/m K 305 300 K 28.3 W/mrad s surεσ −′′ = − = × × ⋅ − = 
 
Winter: ( ) ( )4 4 8 2 4 4 4 4 2q T T 0.9 5.67 10 W/m K 305 287 K 95.4 W/mrad s surεσ −′′ = − = × × ⋅ − = 
 
There is a significant difference between winter and summer radiation fluxes, and the chilled 
condition is attributable to the effect of the colder walls on radiation. 
 
COMMENTS: For a representative surface area of A = 1.5 m2, the heat losses are qconv = 
36 W, qrad(summer) = 42.5 W and qrad(winter) = 143.1 W. The winter time radiation loss is 
significant and if maintained over a 24 h period would amount to 2,950 kcal. 
 
PROBLEM 1.25
KNOWN: Diameter and emissivity of spherical interplanetary probe. Power dissipation
within probe.
FIND: Probe surface temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible radiation incident on the probe.
ANALYSIS: Conservation of energy dictates a balance between energy generation within the
probe and radiation emission from the probe surface. Hence, at any instant
-E + E = 0out g� �
ε σA T Es s
4
g= �
E
T
D
1/ 4
g
s 2επ σ
 
=    
�
( )
150WT
0.8 0.5 m 5.67 10
1/ 4
s 2 8 2 4
 W/m Kπ
  =  × − ⋅
T K.s = 254 7. <
COMMENTS: Incident radiation, as, for example, from the sun, would increase the surface
temperature.
PROBLEM 1.26 
 
KNOWN: Spherical shaped instrumentation package with prescribed surface emissivity within a 
large space-simulation chamber having walls at 77 K. 
 
FIND: Acceptable power dissipation for operating the package surface temperature in the range Ts = 
40 to 85°C. Show graphically the effect of emissivity variations for 0.2 and 0.3. 
 
SCHEMATIC: 
 
 
ASSUMPTIONS: (1) Uniform surface temperature, (2) Chamber walls are large compared to the 
spherical package, and (3) Steady-state conditions. 
 
ANALYSIS: From an overall energy balance on the package, the internal power dissipation Pe will 
be transferred by radiation exchange between the package and the chamber walls. From Eq. 1.7, 
 ( )4 4rad e s s surq P A T Tε σ= = − 
For the condition when Ts = 40°C, with As = piD2 the power dissipation will be 
 ( ) ( )48 2 4 4 4eP 0.25 0.10 m 5.67 10 W m K 40 273 77 K 4.3Wpi −  = × × × ⋅ × + − =   < 
Repeating this calculation for the range 40 ≤ Ts ≤ 85°C, we can obtain the power dissipation as a 
function of surface temperature for the ε = 0.25 condition. Similarly, with 0.2 or 0.3, the family of 
curves shown below has been obtained. 
40 50 60 70 80 90
Surface temperature, Ts (C)
2
4
6
8
10
Po
w
er
 d
iss
ip
at
io
n,
 P
e 
(W
)
Surface emissivity, eps = 0.3
eps = 0.25
eps = 0.2
 
 
COMMENTS: (1) As expected, the internal power dissipation increases with increasing emissivity 
and surface temperature. Because the radiation rate equation is non-linear with respect to 
temperature, the power dissipation will likewise not be linear with surface temperature. 
 
(2) What is the maximum power dissipation that is possible if the surface temperature is not to exceed 
85°C? What kind of a coating should be applied to the instrument package in order to approach this 
limiting condition? 
PROBLEM 1.27
KNOWN: Area, emissivity andtemperature of a surface placed in a large, evacuated
chamber of prescribed temperature.
FIND: (a) Rate of surface radiation emission, (b) Net rate of radiation exchange between
surface and chamber walls.
SCHEMATIC:
ASSUMPTIONS: (1) Area of the enclosed surface is much less than that of chamber walls.
ANALYSIS: (a) From Eq. 1.5, the rate at which radiation is emitted by the surface is
qemit E A = A Ts
4
= ⋅ ε σ
( ) ( )q = 0.8 0.5 m 5.67 10 W/m K 150 + 273 K 42 -8 2 4emit  × ⋅  
q = 726 W.emit <
(b) From Eq. 1.7, the net rate at which radiation is transferred from the surface to the chamber
walls is
( )q = A T T4 4s surε σ −
( ) ( ) ( )q = 0.8 0.5 m 5.67 10 W/m K 423K - 298K4 42 -8 2 4  × ⋅   
q = 547 W. <
COMMENTS: The foregoing result gives the net heat loss from the surface which occurs at
the instant the surface is placed in the chamber. The surface would, of course, cool due to this
heat loss and its temperature, as well as the heat loss, would decrease with increasing time.
Steady-state conditions would eventually be achieved when the temperature of the surface
reached that of the surroundings.
PROBLEM 1.28 
 
KNOWN: Length, diameter, surface temperature and emissivity of steam line. Temperature 
and convection coefficient associated with ambient air. Efficiency and fuel cost for gas fired 
furnace. 
 
FIND: (a) Rate of heat loss, (b) Annual cost of heat loss. 
 
SCHEMATIC: 
 
 
 
 
ASSUMPTIONS: (1) Steam line operates continuously throughout year, (2) Net radiation 
transfer is between small surface (steam line) and large enclosure (plant walls). 
 
ANALYSIS: (a) From Eqs. (1.3a) and (1.7), the heat loss is 
 
 ( ) ( )4 4q q q A h T T T Tconv rad s s surεσ = + = − + −∞   
 
where ( ) 2A DL 0.1m 25m 7.85m .pi pi= = × = 
 
Hence, 
 
( ) ( )2 2 8 2 4 4 4 4q 7.85m 10 W/m K 150 25 K 0.8 5.67 10 W/m K 423 298 K− = ⋅ − + × × ⋅ −   
 
 ( ) ( )2 2q 7.85m 1, 250 1,095 w/m 9813 8592 W 18, 405 W= + = + = < 
 
(b) The annual energy loss is 
 
 
11E qt 18, 405 W 3600 s/h 24h/d 365 d/y 5.80 10 J= = × × × = × 
 
With a furnace energy consumption of 11E E/ 6.45 10 J,f fη= = × the annual cost of the loss 
is 
 
 
5C C E 0.01 $/MJ 6.45 10 MJ $6450g f= = × × = < 
 
COMMENTS: The heat loss and related costs are unacceptable and should be reduced by 
insulating the steam line. 
 
PROBLEM 1.29 
 
KNOWN: Exact and approximate expressions for the linearized radiation coefficient, hr and hra, 
respectively. 
 
FIND: (a) Comparison of the coefficients with ε = 0.05 and 0.9 and surface temperatures which may 
exceed that of the surroundings (Tsur = 25°C) by 10 to 100°C; also comparison with a free convection 
coefficient correlation, (b) Plot of the relative error (hr - rra)/hr as a function of the furnace temperature 
associated with a workpiece at Ts = 25°C having ε = 0.05, 0.2 or 0.9. 
 
ASSUMPTIONS: (1) Furnace walls are large compared to the workpiece and (2) Steady-state 
conditions. 
 
ANALYSIS: (a) The linearized radiation coefficient, Eq. 1.9, follows from the radiation exchange 
rate equation, 
 ( )( )2 2r s sur s surh T T T Tεσ= + + 
If Ts ≈ Tsur, the coefficient may be approximated by the simpler expression 
 ( )3r,a s surh 4 T T T T 2εσ= = + 
For the condition of ε = 0.05, Ts = Tsur + 10 = 35°C = 308 K and Tsur = 25°C = 298 K, find that 
 ( )( )8 2 4 2 2 3 2rh 0.05 5.67 10 W m K 308 298 308 298 K 0.32 W m K−= × × ⋅ + + = ⋅ < 
 ( )( )38 2 4 3 2r,ah 4 0.05 5.67 10 W m K 308 298 2 K 0.32 W m K−= × × × ⋅ + = ⋅ < 
The free convection coefficient with Ts = 35°C and T∞ = Tsur = 25°C, find that 
 ( ) ( )1/3 1/31/3 2sh 0.98 T 0.98 T T 0.98 308 298 2.1W m K∞= ∆ = − = − = ⋅ < 
For the range Ts - Tsur = 10 to 100°C with ε = 0.05 and 0.9, the results for the coefficients are 
tabulated below. For this range of surface and surroundings temperatures, the radiation and free 
convection coefficients are of comparable magnitude for moderate values of the emissivity, say ε > 
0.2. The approximate expression for the linearized radiation coefficient is valid within 2% for these 
conditions. 
 
(b) The above expressions for the radiation coefficients, hr and hr,a, are used for the workpiece at Ts = 
25°C placed inside a furnace with walls which may vary from 100 to 1000°C. The relative error, (hr - 
hra)/hr, will be independent of the surface emissivity and is plotted as a function of Tsur. For Tsur > 
150°C, the approximate expression provides estimates which are in error more than 5%. The 
approximate expression should be used with caution, and only for surface and surrounding 
temperature differences of 50 to 100°C. 
 
 
Coefficients (W/m2⋅K) 
Ts (°C) ε hr hr,a h 
35 0.05 0.32 0.32 2.1 
 0.9 5.7 5.7 
135 0.05 0.51 0.50 4.7 
 0.9 9.2 9.0 
 
 
 
 
100 300 500 700 900
Surroundings temperature, Tsur (C)
0
10
20
30
R
el
at
iv
e 
er
ro
r, 
(hr
-hr
a)/
hr*
10
0 (
%)
 
 
 
PROBLEM 1.30
KNOWN: Chip width, temperature, and heat loss by convection in air. Chip emissivity and
temperature of large surroundings.
FIND: Increase in chip power due to radiation.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Radiation exchange between small surface
and large enclosure.
ANALYSIS: Heat transfer from the chip due to net radiation exchange with the surroundings
is
( )q = W T - T2 4 4surrad ε σ
( ) ( )q = 0.9 0.005 m 5.67 10 W/m K 358 - 288 K2 8 2 4 4 4 4rad × ⋅−
qrad = 0.0122 W.
The percent increase in chip power is therefore
∆P
P
qrad
qconv
 W
0.350 W
× = × = × =100 100 0 0122 100 35%.. . <
COMMENTS: For the prescribed conditions, radiation effects are small. Relative to
convection, the effect of radiation would increase with increasing chip temperature and
decreasing convection coefficient.
PROBLEM 1.31 
 
KNOWN: Width, surface emissivity and maximum allowable temperature of an electronic chip. 
Temperature of air and surroundings. Convection coefficient. 
 
FIND: (a) Maximum power dissipation for free convection with h(W/m2⋅K) = 4.2(T - T∞)1/4, (b) 
Maximum power dissipation for forced convection with h = 250 W/m2⋅K. 
 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) Steady-state conditions, (2) Radiation exchange between a small surface and a 
large enclosure, (3) Negligible heat transfer from sides of chip or from back of chip by conduction 
through the substrate. 
 
ANALYSIS: Subject to the foregoing assumptions, electric power dissipation by the chip must be 
balanced by convection and radiation heat transfer from the chip. Hence, from Eq. (1.10), 
 ( ) ( )4 4P q q hA T T A T Telec conv rad s s surε σ= + = − + −∞ 
where ( )22 4 2A L 0.015m 2.25 10 m .−= = = × 
 
(a) If heat transfer is by natural convection, 
 ( ) ( )( )5 / 4 5 / 42 5/4 4 2q C A T T 4.2 W/m K 2.25 10 m 60K 0.158 Wconv s −= − = ⋅ × =∞ 
 ( ) ( )4 2 8 2 4 4 4 4q 0.60 2.25 10 m 5.67 10 W/m K 358 298 K 0.065 Wrad − −= × × ⋅ − = 
 P 0.158 W 0.065 W 0.223 Welec = + = < 
(b) If heat transfer is by forced convection, 
 
 ( ) ( )( )2 4 2q hA T T 250 W/m K 2.25 10 m 60K 3.375 Wconv s −= − = ⋅ × =∞ 
 
 P 3.375 W 0.065 W 3.44 Welec = + = < 
 
COMMENTS: Clearly, radiation and natural convection are inefficient mechanisms for transferring 
heat from the chip. For Ts = 85°C and T∞ = 25°C, the natural convection coefficient is 11.7 W/m
2
⋅K. 
Even for forced convection with h = 250 W/m2⋅K, the power dissipation is well below that associated 
with many of today’s processors. To provide acceptable cooling, it is often necessary to attach the 
chip to a highly conducting substrate and to thereby provide an additional heat transfer mechanism 
due to conduction from the backsurface. 
PROBLEM 1.32
KNOWN: Vacuum enclosure maintained at 77 K by liquid nitrogen shroud while baseplate is
maintained at 300 K by an electrical heater.
FIND: (a) Electrical power required to maintain baseplate, (b) Liquid nitrogen consumption rate, (c)
Effect on consumption rate if aluminum foil (εp = 0.09) is bonded to baseplate surface.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) No heat losses from backside of heater or sides of
plate, (3) Vacuum enclosure large compared to baseplate, (4) Enclosure is evacuated with negligible
convection, (5) Liquid nitrogen (LN2) is heated only by heat transfer to the shroud, and (6) Foil is
intimately bonded to baseplate.
PROPERTIES: Heat of vaporization of liquid nitrogen (given): 125 kJ/kg.
ANALYSIS: (a) From an energy balance on the baseplate,
� �E - E = 0 q - q = 0in out elec rad
and using Eq. 1.7 for radiative exchange between the baseplate and shroud,
( )pq = A T - T .4 4p pelec shε σ
Substituting numerical values, with ( )A = D / 4 ,2p pπ find
( ) ( )q = 0.25 0.3 m / 4 5.67 10 W/m K 300 - 77 K 8.1 W.2 8 2 4 4 4 4elec π  × ⋅ =   − <
(b) From an energy balance on the enclosure, radiative transfer heats the liquid nitrogen stream
causing evaporation,
� � �E - E = 0 q - m h = 0in out rad LN2 fg
where �mLN2 is the liquid nitrogen consumption rate. Hence,
� /mLN2 = qrad hfg = 8.1 W / 125 kJ / kg = 6.48 10
-5
 kg / s = 0.23 kg / h.× <
(c) If aluminum foil (εp = 0.09) were bonded to the upper surface of the baseplate,( ) ( )q = q / = 8.1 W 0.09/0.25 = 2.9 Wprad,foil rad fε ε
and the liquid nitrogen consumption rate would be reduced by
(0.25 - 0.09)/0.25 = 64% to 0.083 kg/h. <
PROBLEM 1.33 
 
KNOWN: Width, input power and efficiency of a transmission. Temperature and convection 
coefficient for air flow over the casing. Emissivity of casing and temperature of surroundings. 
 
FIND: Surface temperature of casing. 
 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) Steady state, (2) Uniform convection coefficient and surface temperature, (3) 
Radiation exchange with large surroundings. 
 
ANALYSIS: Heat transfer from the case must balance heat dissipation in the transmission, which 
may be expressed as q = Pi – Po = Pi (1 - η) = 150 hp × 746 W/hp × 0.07 = 7833 W. Heat transfer 
from the case is by convection and radiation, in which case 
 
 ( ) ( )4 4s s s surq A h T T T Tεσ∞ = − + −   
 
where As = 6 W
2
. Hence, 
 
( ) ( ) ( )2 2 2 4 4 4 4s s87833 W 6 0.30 m 200 W / m K T 303K 0.8 5.67 10 W / m K T 303 K−= ⋅ − + × × ⋅ −    
 
A trial-and-error solution yields 
 
 sT 373K 100 C≈ = ° < 
 
COMMENTS: (1) For Ts ≈ 373 K, qconv ≈ 7,560 W and qrad ≈ 270 W, in which case heat transfer is 
dominated by convection, (2) If radiation is neglected, the corresponding surface temperature is Ts = 
102.5°C. 
 
PROBLEM 1.34 
 
KNOWN: Resistor connected to a battery operating at a prescribed temperature in air. 
 
FIND: (a) Considering the resistor as the system, determine corresponding values for ( )inE W� , 
( )gE W� , ( )outE W� and ( )stE W� . If a control surface is placed about the entire system, determine 
the values for inE� , gE� , outE� , and stE� . (b) Determine the volumetric heat generation rate within 
the resistor, q� (W/m3), (c) Neglecting radiation from the resistor, determine the convection 
coefficient. 
 
SCHEMATIC: 
 
 
ASSUMPTIONS: (1) Electrical power is dissipated uniformly within the resistor, (2) Temperature 
of the resistor is uniform, (3) Negligible electrical power dissipated in the lead wires, (4) Negligible 
radiation exchange between the resistor and the surroundings, (5) No heat transfer occurs from the 
battery, (5) Steady-state conditions. 
 
ANALYSIS: (a) Referring to Section 1.3.1, the conservation of energy requirement for a control 
volume at an instant of time, Eq 1.11a, is 
 
 in g out stE E E E+ − =� � � � 
 
where in outE , E� � correspond to surface inflow and outflow processes, respectively. The energy 
generation term gE� is associated with conversion of some other energy form (chemical, electrical, 
electromagnetic or nuclear) to thermal energy. The energy storage term stE� is associated with 
changes in the internal, kinetic and/or potential energies of the matter in the control volume. gE� , 
stE� are volumetric phenomena. The electrical power delivered by the battery is P = VI = 24V×6A = 
144 W. 
 
Control volume: Resistor. 
 
 in outE 0 E 144 W= =� � 
< 
 g stE 144 W E 0= =� � 
 
 
The gE� term is due to conversion of electrical energy to thermal energy. The term outE� is due to 
convection from the resistor surface to the air. 
Continued... 
 
 
PROBLEM 1.34 (Cont.) 
 
 
 
Control volume: Battery-Resistor System. 
 
 in outE 0 E 144 W= =� � < 
 g stE 0 E 144 W= = −� � 
 
 
 
 
 
The stE� term represents the decrease in the chemical energy within the battery. The conversion of 
chemical energy to electrical energy and its subsequent conversion to thermal energy are processes 
internal to the system which are not associated with stE� or gE� . The outE� term is due to convection 
from the resistor surface to the air. 
 
(b) From the energy balance on the resistor with volume, ∀ = (piD2/4)L, 
 
 ( )( )2 5 3gE q 144 W q 0.06 m / 4 0.25m q 2.04 10 W mpi= ∀ = × = ×� � � � < 
 
(c) From the energy balance on the resistor and Newton's law of cooling with As = piDL + 2(piD2/4), 
 
 ( )out cv s sE q hA T T∞= = −� 
 
 ( ) ( )2 2144 W h 0.06 m 0.25m 2 0.06 m 4 95 25 Cpi pi = × × + × −   $ 
 
 [ ] ( )2144 W h 0.0471 0.0057 m 95 25 C= + − $ 
 
2h 39.0 W m K= ⋅ < 
 
COMMENTS: (1) In using the conservation of energy requirement, Eq. 1.11a, it is important to 
recognize that inE� and outE� will always represent surface processes and gE� and stE� , volumetric 
processes. The generation term gE� is associated with a conversion process from some form of 
energy to thermal energy. The storage term stE� represents the rate of change of internal energy. 
 
(2) From Table 1.1 and the magnitude of the convection coefficient determined from part (c), we 
conclude that the resistor is experiencing forced, rather than free, convection. 
 
PROBLEM 1.35 
 
KNOWN: Thickness and initial temperature of an aluminum plate whose thermal environment is 
changed. 
 
FIND: (a) Initial rate of temperature change, (b) Steady-state temperature of plate, (c) Effect of 
emissivity and absorptivity on steady-state temperature. 
 
SCHEMATIC: 
 
 
ASSUMPTIONS: (1) Negligible end effects, (2) Uniform plate temperature at any instant, (3) 
Constant properties, (4) Adiabatic bottom surface, (5) Negligible radiation from surroundings, (6) No 
internal heat generation. 
 
ANALYSIS: (a) Applying an energy balance, Eq. 1.11a, at an instant of time to a control volume 
about the plate, in out stE E E− =� � � , it follows for a unit surface area. 
 ( ) ( ) ( ) ( )( ) ( ) ( )2 2 2 2S S convG 1m E 1m q 1m d dt McT 1m L c dT dtα ρ′′− − = = × . 
Rearranging and substituting from Eqs. 1.3 and 1.5, we obtain 
 ( ) ( )4S S i idT dt 1 Lc G T h T Tρ α εσ ∞= − − −   . 
 ( ) 13dT dt 2700 kg m 0.004 m 900 J kg K −= × × ⋅ × 
 ( ) ( )42 8 2 4 20.8 900 W m 0.25 5.67 10 W m K 298 K 20 W m K 25 20 C−× − × × ⋅ − ⋅ −   
$
 
 dT dt 0.052 C s= $ . < 
(b) Under steady-state conditions, �Est = 0, and the energy balance reduces to 
 ( )4S SG T h T Tα εσ ∞= + − (2) 
 ( )2 8 2 4 4 20.8 900 W m 0.25 5.67 10 W m K T 20 W m K T 293K−× = × × ⋅ × + ⋅ − 
The solution yields T = 321.4 K = 48.4°C. < 
(c) Using the IHT First LawModel for an Isothermal Plane Wall, parametric calculations yield the 
following results. 
0 0.2 0.4 0.6 0.8 1
Coating emissivity, eps
20
30
40
50
60
70
Pl
at
e 
te
m
pe
ra
tu
re
, T
 (C
)
Solar absorptivity, alphaS = 1
alphaS = 0.8
alphaS = 0.5
 
COMMENTS: The surface radiative properties have a significant effect on the plate temperature, 
which decreases with increasing ε and decreasing αS. If a low temperature is desired, the plate 
coating should be characterized by a large value of ε/αS. The temperature also decreases with 
increasing h. 
PROBLEM 1.36
KNOWN: Surface area of electronic package and power dissipation by the electronics.
Surface emissivity and absorptivity to solar radiation. Solar flux.
FIND: Surface temperature without and with incident solar radiation.
SCHEMATIC:
ASSUMPTIONS: Steady-state conditions.
ANALYSIS: Applying conservation of energy to a control surface about the compartment, at
any instant
� � �E - E + E = 0.in out g
It follows that, with the solar input,
A q A E P=0
A q A T P=0
A q P
T .
A
s sS S
4
s s sS S
1/ 4
sS S
s
s
α
α εσ
α
εσ
′′
− +
′′
− +
′′ +
=   
In the shade ( )q 0 ,S′′ =
1000 WT 364 K.
1 m 1 5.67 10 W/m K
1/ 4
s 2 8 2 4
 
= = 
× × × ⋅ − <
In the sun,
0.25 1 m 750 W/m 1000 WT 380 K.
1 m 1 5.67 10 W/m K
1/ 42 2
s 2 8 2 4
 × × + = = × × × ⋅ −
<
COMMENTS: In orbit, the space station would be continuously cycling between shade and
sunshine, and a steady-state condition would not exist.
PROBLEM 1.37 
 
KNOWN: Daily hot water consumption for a family of four and temperatures associated with ground 
water and water storage tank. Unit cost of electric power. Heat pump COP. 
 
FIND: Annual heating requirement and costs associated with using electric resistance heating or a 
heat pump. 
 
SCHEMATIC: 
 
 
ASSUMPTIONS: (1) Process may be modelled as one involving heat addition in a closed system, 
(2) Properties of water are constant. 
 
PROPERTIES: Table A-6, Water ( aveT = 308 K): ρ = 1fv
−
 = 993 kg/m3, p,fc = 4.178 kJ/kg⋅K. 
 
ANALYSIS: From Eq. 1.11c, the daily heating requirement is daily tQ U Mc T= ∆ = ∆ 
( )f iVc T Tρ= − . With V = 100 gal/264.17 gal/m3 = 0.379 m3, 
 
 ( ) ( )3 3dailyQ 993kg / m 0.379m 4.178kJ/kg K 40 C 62,900 kJ= ⋅ =$ 
 
The annual heating requirement is then, ( ) 7annualQ 365days 62,900 kJ/day 2.30 10 kJ= = × , or, 
with 1 kWh = 1 kJ/s (3600 s) = 3600 kJ, 
 
 annualQ 6380kWh= < 
 
With electric resistance heating, annual elecQ Q= and the associated cost, C, is 
 
 ( )C 6380kWh $0.08/kWh $510= = < 
 
If a heat pump is used, ( )annual elecQ COP W .= Hence, 
 
 ( ) ( )elec annualW Q / COP 6380kWh/ 3 2130kWh= = = 
 
The corresponding cost is 
 
 ( )C 2130 kWh $0.08/kWh $170= = < 
 
COMMENTS: Although annual operating costs are significantly lower for a heat pump, 
corresponding capital costs are much higher. The feasibility of this approach depends on other factors 
such as geography and seasonal variations in COP, as well as the time value of money. 
PROBLEM 1.38 
 
KNOWN: Initial temperature of water and tank volume. Power dissipation, emissivity, 
length and diameter of submerged heaters. Expressions for convection coefficient associated 
with natural convection in water and air. 
 
FIND: (a) Time to raise temperature of water to prescribed value, (b) Heater temperature 
shortly after activation and at conclusion of process, (c) Heater temperature if activated in air. 
 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) Negligible heat loss from tank to surroundings, (2) Water is well-
mixed (at a uniform, but time varying temperature) during heating, (3) Negligible changes in 
thermal energy storage for heaters, (4) Constant properties, (5) Surroundings afforded by tank 
wall are large relative to heaters. 
 
ANALYSIS: (a) Application of conservation of energy to a closed system (the water) at an 
instant, Eq. (1.11d), yields 
 
 
dU dT dTMc c q 3q1dt dt dt
ρ= = ∀ = = 
 
Hence, ( )
Tft dt c/3q dT10 Ti
ρ= ∀∫ ∫ 
 
 
( ) ( )3 3 3990 kg/m 10gal 3.79 10 m / gal 4180J/kg Kt 335 295 K 4180 s
3 500 W
−× × ⋅
= − =
×
 < 
 
(b) From Eq. (1.3a), the heat rate by convection from each heater is 
 
 ( ) ( ) ( )4 / 3q Aq Ah T T DL 370 T T1 1 s spi′′= = − = − 
 
Hence, 
( )
3/ 43/ 4q 500 W1T T T T 24 Ks 2 4/3370 DL 370 W/m K 0.025 m 0.250 mpi pi
  
= + = + = +   
⋅ × × ×  
 
With water temperatures of Ti ≈ 295 K and Tf = 335 K shortly after the start of heating and at 
the end of heating, respectively, 
 
 Ts,i = 319 K Ts,f = 359 K < 
 
 Continued ….. 
PROBLEM 1.38 (Continued) 
 
(c) From Eq. (1.10), the heat rate in air is 
 
 ( ) ( )4 / 3 4 4q DL 0.70 T T T T1 s s surpi εσ = − + −∞   
 
Substituting the prescribed values of q1, D, L, T∞ = Tsur and ε, an iterative solution yields 
 
 Ts = 830 K < 
 
COMMENTS: In part (c) it is presumed that the heater can be operated at Ts = 830 K 
without experiencing burnout. The much larger value of Ts for air is due to the smaller 
convection coefficient. However, with qconv and qrad equal to 59 W and 441 W, respectively, 
a significant portion of the heat dissipation is effected by radiation. 
 
PROBLEM 1.39 
 
KNOWN: Power consumption, diameter, and inlet and discharge temperatures of a hair 
dryer. 
 
FIND: (a) Volumetric flow rate and discharge velocity of heated air, (b) Heat loss from case. 
 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) Steady-state, (2) Constant air properties, (3) Negligible potential and 
kinetic energy changes of air flow, (4) Negligible work done by fan, (5) Negligible heat 
transfer from casing of dryer to ambient air (Part (a)), (6) Radiation exchange between a small 
surface and a large enclosure (Part (b)). 
 
ANALYSIS: (a) For a control surface about the air flow passage through the dryer, 
conservation of energy for an open system reduces to 
 
 ( ) ( )m u pv m u pv q 0i o+ − + + =� � 
 
where u + pv = i and q = Pelec. Hence, with ( ) ( )m i i mc T T ,i o p i o− = −� � 
 
 ( )mc T T Pp o i elec− =� 
 
 ( ) ( )
P 500 Welecm 0.0199 kg/s
c T Tp o i 1007 J/kg K 25 C
= = =
−
⋅
�
$
 
 
 
m 0.0199 kg/s 30.0181 m / s31.10 kg/mρ
∀ = = =�� < 
 
 
( )
34 4 0.0181 m / sV 4.7 m/so 2 2Ac D 0.07 mpi pi
∀ ∀ ×
= = = =
� �
 < 
 
(b) Heat transfer from the casing is by convection and radiation, and from Eq. (1.10) 
 
 ( ) ( )4 4q hA T T A T Ts s s s surε σ= − + −∞ 
 Continued ….. 
PROBLEM 1.39 (Continued) 
 
where ( ) 2A DL 0.07 m 0.15 m 0.033 m .s pi pi= = × = Hence, 
 
( )( ) ( )2 2 2 8 2 4 4 4 4q 4W/m K 0.033 m 20 C 0.8 0.033 m 5.67 10 W/m K 313 293 K−= ⋅ + × × × ⋅ −$
 
 q 2.64 W 3.33 W 5.97 W= + = < 
 
The heat loss is much less than the electrical power, and the assumption of negligible heat loss 
is justified. 
 
COMMENTS: Although the mass flow rate is invariant, the volumetric flow rate increases 
as the air is heated in its passage through the dryer, causing a reduction in the density. 
However, for the prescribed temperature rise, the change in ρ, and hence the effect on ,∀� is 
small. 
 
 
PROBLEM 1.40 
 
KNOWN: Speed, width, thickness and initial and final temperatures of 304 stainless steel in an 
annealing process. Dimensions of annealing oven and temperature, emissivity and convection 
coefficient of surfaces exposed to ambient air and large surroundings of equivalent temperatures. 
Thickness of pad on which oven rests and pad surface temperatures. 
 
FIND: Oven operating power. 
 
SCHEMATIC:ASSUMPTIONS: (1) steady-state, (2) Constant properties, (3) Negligible changes in kinetic and 
potential energy. 
 
PROPERTIES: Table A.1, St.St.304 ( )( )i oT T T /2 775K= + = : ρ = 7900 kg/m3, cp = 578 
J/kg⋅K; Table A.3, Concrete, T = 300 K: kc = 1.4 W/m⋅K. 
 
ANALYSIS: The rate of energy addition to the oven must balance the rate of energy transfer to the 
steel sheet and the rate of heat loss from the oven. With in outE E−� � = 0, it follows that 
 
 ( )elec i oP m u u q 0+ − − =� 
 
where heat is transferred from the oven. With ( )s s sm V W tρ=� , ( ) ( )i o p i ou u c T T− = − , and 
( ) ( ) ( )4 4o o o o o o s s s surq 2H L 2H W W L h T T T Tε σ∞ = + + × − + −   ( )( )c o o s b ck W L T T /t+ − , 
it follows that 
 ( ) ( ) ( )elec s s s p o i o o o o o oP V W t c T T 2H L 2H W W Lρ= − + + + × 
 ( ) ( ) ( )( )4 4s o s s sur c o o s b ch T T T T k W L T T /tε σ − + − + −   
 ( ) ( )3elecP 7900kg/m 0.01m/s 2 m 0.008m 578J/kg K 1250 300 K= × × ⋅ − 
 ( ) ( )22 2m 25m 2 2m 2.4m 2.4m 25m [10W/m K 350 300 K+ × × + × × + × ⋅ − 
( ) ( )( )8 2 4 4 4 40.8 5.67 10 W/m K 350 300 K ] 1.4W/m K 2.4m 25m 350 300 K/0.5m−+ × × ⋅ − + ⋅ × − 
 
Continued.…. 
 
PROBLEM 1.40 (Cont.) 
 
 ( )2 2elecP 694,000W 169.6m 500 313 W/m 8400W= + + + 
 
 ( )694,000 84,800 53,100 8400 W 840kW= + + + = < 
 
COMMENTS: Of the total energy input, 83% is transferred to the steel while approximately 10%, 
6% and 1% are lost by convection, radiation and conduction from the oven. The convection and 
radiation losses can both be reduced by adding insulation to the side and top surfaces, which would 
reduce the corresponding value of Ts. 
PROBLEM 1.41 
 
KNOWN: Hot plate-type wafer thermal processing tool based upon heat transfer modes by 
conduction through gas within the gap and by radiation exchange across gap. 
 
FIND: (a) Radiative and conduction heat fluxes across gap for specified hot plate and wafer 
temperatures and gap separation; initial time rate of change in wafer temperature for each mode, and 
(b) heat fluxes and initial temperature-time change for gap separations of 0.2, 0.5 and 1.0 mm for hot 
plate temperatures 300 < Th < 1300°C. Comment on the relative importance of the modes and the 
influence of the gap distance. Under what conditions could a wafer be heated to 900°C in less than 10 
seconds? 
 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) Steady-state conditions for flux calculations, (2) Diameter of hot plate and 
wafer much larger than gap spacing, approximating plane, infinite planes, (3) One-dimensional 
conduction through gas, (4) Hot plate and wafer are blackbodies, (5) Negligible heat losses from wafer 
backside, and (6) Wafer temperature is uniform at the onset of heating. 
 
PROPERTIES: Wafer: ρ = 2700 kg/m3, c = 875 J/kg⋅K; Gas in gap: k = 0.0436 W/m⋅K. 
 
ANALYSIS: (a) The radiative heat flux between the hot plate and wafer for Th = 600°C and Tw = 
20° C follows from the rate equation, 
 ( ) ( ) ( )( )4 44 4 8 2 4 4 2rad h wq T T 5.67 10 W / m K 600 273 20 273 K 32.5 kW / mσ −′′ = − × ⋅ + − + == < 
 
The conduction heat flux through the gas in the gap with L = 0.2 mm follows from Fourier’s law, 
 
 
( ) 2h w
cond
600 20 KT Tq k 0.0436 W / m K 126 kW / m
L 0.0002 m
−
−
′′ = = ⋅ = < 
 
The initial time rate of change of the wafer can be determined from an energy balance on the wafer at 
the instant of time the heating process begins, 
 
 
w
in out st st
i
dTE E E E cd
dt
ρ  ′′ ′′ ′′ ′′− = =   � � � � 
 
where outE 0′′ =� and in rad condE q or q .′′ ′′ ′′=� Substituting numerical values, find 
 
 
3 2
w rad
3i,rad
dT q 32.5 10 W / m 17.6 K / s
dt cd 2700 kg / m 875 J / kg K 0.00078 mρ
′′ ×
= = = × ⋅ × < 
 
 
w cond
i,cond
dT q 68.4 K / s
dt cdρ
′′
= = < 
 Continued ….. 
PROBLEM 1.41 (Cont.) 
 
(b) Using the foregoing equations, the heat fluxes and initial rate of temperature change for each mode 
can be calculated for selected gap separations L and range of hot plate temperatures Th with Tw = 
20°C. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
In the left-hand graph, the conduction heat flux increases linearly with Th and inversely with L as 
expected. The radiative heat flux is independent of L and highly non-linear with Th, but does not 
approach that for the highest conduction heat rate until Th approaches 1200°C. 
 
The general trends for the initial temperature-time change, (dTw/dt)i, follow those for the heat fluxes. 
To reach 900°C in 10 s requires an average temperature-time change rate of 90 K/s. Recognizing that 
(dTw/dt) will decrease with increasing Tw, this rate could be met only with a very high Th and the 
smallest L. 
 
300 500 700 900 1100 1300
Hot plate temperature, Th (C)
0
100
200
300
400
H
e
a
t f
lu
x 
(kW
/m
^
2)
q''rad
q''cond, L = 1.0 mm
q''cond, L = 0.5 mm
q''cond, L = 0.2 mm
300 500 700 900 1100 1300
Hot plate temperature, Th (C)
0
50
100
150
200
In
iti
a
l r
a
te
 
o
f c
ha
ng
e,
 
dT
w
/d
t (K
.
s
^
-
1)
q''rad
q''cond, L = 1.0 mm
q''cond, L = 0.5 mm
q''cond, L = 0.2 mm
PROBLEM 1.42 
 
KNOWN: Silicon wafer, radiantly heated by lamps, experiencing an annealing process with known 
backside temperature. 
 
FIND: Whether temperature difference across the wafer thickness is less than 2°C in order to avoid 
damaging the wafer. 
 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) Steady-state conditions, (2) One-dimensional conduction in wafer, (3) 
Radiation exchange between upper surface of wafer and surroundings is between a small object and a 
large enclosure, and (4) Vacuum condition in chamber, no convection. 
 
PROPERTIES: Wafer: k = 30 W/m⋅K, 0.65.ε α= =
"
 
 
ANALYSIS: Perform a surface energy balance on the upper surface of the wafer to determine 
w,uT . The processes include the absorbed radiant flux from the lamps, radiation exchange with the 
chamber walls, and conduction through the wafer. 
 
 in outE E 0′′ ′′− =� � 
 
 s rad cdq q q 0α ′′ ′′ ′′− − =" 
 
 ( ) w,u w,4 4s w,u sur T Tq T T k 0Lα εσ −′′ − − − ="" 
 
 ( )( )45 2 8 2 4 4 4w,u0.65 3.0 10 W / m 0.65 5.67 10 W / m K T 27 273 K−× × − × × ⋅ − + 
 
 ( )w,u30W / m K T 997 273 K / 0.00078 m 0 − ⋅ − + =  
 
 w,uT 1273K 1000 C= = ° < 
 
COMMENTS: (1) The temperature difference for this steady-state operating condition, 
w,u w,lT T ,− is larger than 2°C. Warping of the wafer and inducing slip planes in the crystal structure 
could occur. 
 
(2) The radiation exchange rate equation requires that temperature must be expressed in kelvin units. 
Why is it permissible to use kelvin or Celsius temperature units in the conduction rate equation? 
 
(3) Note how the surface energy balance, Eq. 1.12, is represented schematically. It is essential to 
show the control surfaces, and then identify the rate processes associated with the surfaces. Make 
sure the directions (in or out) of the process are consistent with the energy balance equation. 
PROBLEM 1.43 
 
KNOWN: Silicon wafer positioned in furnace with top and bottom surfaces exposed to hot and cool 
zones, respectively. 
 
FIND: (a) Initial rate of change of the wafer temperature corresponding to the wafer temperature 
w,iT 300 K,= and (b) Steady-state temperature reached if the wafer remains in this position. How 
significant is convection for this situation? Sketch how you’d expect the wafer temperature to vary as 
a function of vertical distance. 
 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) Wafer temperature is uniform, (2) Transient conditionswhen wafer is initially 
positioned, (3) Hot and cool zones have uniform temperatures, (3) Radiation exchange is between 
small surface (wafer) and large enclosure (chamber, hot or cold zone), and (4) Negligible heat losses 
from wafer to mounting pin holder. 
 
ANALYSIS: The energy balance on the wafer illustrated in the schematic above includes convection 
from the upper (u) and lower (l) surfaces with the ambient gas, radiation exchange with the hot- and 
cool-zone (chamber) surroundings, and the rate of energy storage term for the transient condition. 
 in out stE E E′′ ′′ ′′− =� � � 
 
 
w
rad,h rad,c cv,u cv,l
d Tq q q q cd
dt
ρ′′ ′′ ′′ ′′+ − − = 
 
 ( ) ( ) ( ) ( )4 4 4 4 ww sur,c w u w l wsur,h d TT T T T h T T h T T cd dtεσ εσ ρ∞ ∞− + − − − − − = 
 
(a) For the initial condition, the time rate of temperature change of the wafer is determined using the 
energy balance above with w w,iT T 300 K,= = 
 ( ) ( )8 2 4 4 4 8 2 4 4 4 440.65 5.67 10 W / m K 1500 300 K 0.65 5.67 10 W / m K 330 300 K− −× × ⋅ − + × × ⋅ − 
 
 ( ) ( )2 28 W / m K 300 700 K 4 W / m K 300 700 K− ⋅ − − ⋅ − = 
 
 
32700 kg / m 875J / kg K× ⋅ ( )w i0.00078 m d T / dt× 
 
 ( )w id T / dt 104 K / s= < 
 
(b) For the steady-state condition, the energy storage term is zero, and the energy balance can be 
solved for the steady-state wafer temperature, w w,ssT T .= 
 
 Continued ….. 
PROBLEM 1.43 (Cont.) 
 
( ) ( )4 4 4 4 4 4w,ss w,ss0.65 1500 T K 0.65 330 T Kσ σ− + − 
 
 ( ) ( )2 2w,ss w,ss8 W / m K T 700 K 4 W / m K T 700 K 0− ⋅ − − ⋅ − = 
 w,ssT 1251 K= < 
 
To determine the relative importance of the convection processes, re-solve the energy balance above 
ignoring those processes to find ( )w w,ssid T / dt 101 K / s and T 1262 K.= = We conclude that the 
radiation exchange processes control the initial time rate of temperature change and the steady-state 
temperature. 
 
If the wafer were elevated above the present operating position, its temperature would increase, since 
the lower surface would begin to experience radiant exchange with progressively more of the hot zone 
chamber. Conversely, by lowering the wafer, the upper surface would experience less radiant 
exchange with the hot zone chamber, and its temperature would decrease. The temperature-distance 
trend might appear as shown in the sketch. 
 
 
 
 
 
PROBLEM 1.44
KNOWN: Radial distribution of heat dissipation in a cylindrical container of radioactive
wastes. Surface convection conditions.
FIND: Total energy generation rate and surface temperature.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Negligible temperature drop across thin
container wall.
ANALYSIS: The rate of energy generation is
( )
( )
oE qdV=q 1- r/r 2 rLdr
E 2 Lq r / 2 r / 4
r 2
g o o0
2 2g o o o
π
π
 
=   
= −
∫ ∫� � �
� �
or per unit length,
�
�
.′ =E q r
2g
o o
2π <
Performing an energy balance for a control surface about the container yields, at an instant,
� �′ − ′ =E Eg out 0
and substituting for the convection heat rate per unit length,
( )( )q r h 2 r T T
2
2
o o
o s
π
π= − ∞
�
T T q r
4hs
o o
= +
∞
�
. <
COMMENTS: The temperature within the radioactive wastes increases with decreasing r
from Ts at ro to a maximum value at the centerline.
PROBLEM 1.45 
 
KNOWN: Rod of prescribed diameter experiencing electrical dissipation from passage of electrical 
current and convection under different air velocity conditions. See Example 1.3. 
 
FIND: Rod temperature as a function of the electrical current for 0 ≤ I ≤ 10 A with convection 
coefficients of 50, 100 and 250 W/m2⋅K. Will variations in the surface emissivity have a significant 
effect on the rod temperature? 
 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform rod temperature, (3) Radiation exchange 
between the outer surface of the rod and the surroundings is between a small surface and large 
enclosure. 
 
ANALYSIS: The energy balance on the rod for steady-state conditions has the form, 
 
 conv rad genq q E′ ′ ′+ = � 
 
 ( ) ( )4 4 2sur eDh T T D T T I Rpi pi εσ∞ ′− + − = 
 
Using this equation in the Workspace of IHT, the rod temperature is calculated and plotted as a 
function of current for selected convection coefficients. 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
COMMENTS: (1) For forced convection over the cylinder, the convection heat transfer coefficient is 
dependent upon air velocity approximately as h ~ V0.6. Hence, to achieve a 5-fold change in the 
convection coefficient (from 50 to 250 W/m2⋅K), the air velocity must be changed by a factor of 
nearly 15. 
 
 Continued ….. 
0 2 4 6 8 1 0
C u rre n t, I (a m p e re s )
0
2 5
5 0
7 5
1 0 0
1 2 5
1 5 0
Ro
d 
te
m
pe
ra
tu
re
, 
T 
(C
)
h = 5 0 W /m ^2 .K
h = 1 0 0 W /m ^2 .K
h = 2 5 0 W /m ^2 .K
PROBLEM 1.45 (Cont.) 
 
(2) For the condition of I = 4 A with h = 50 W/m2⋅K with T = 63.5°C, the convection and radiation 
exchange rates per unit length are, respectively, cv radq 5.7 W / m and q 0.67 W / m.′ ′= = We conclude 
that convection is the dominate heat transfer mode and that changes in surface emissivity could have 
only a minor effect. Will this also be the case if h = 100 or 250 W/m2⋅K? 
 
(3) What would happen to the rod temperature if there was a “loss of coolant” condition where the air 
flow would cease? 
 
(4) The Workspace for the IHT program to calculate the heat losses and perform the parametric 
analysis to generate the graph is shown below. It is good practice to provide commentary with the 
code making your solution logic clear, and to summarize the results. It is also good practice to show 
plots in customary units, that is, the units used to prescribe the problem. As such the graph of the rod 
temperature is shown above with Celsius units, even though the calculations require temperatures in 
kelvins. 
 
 
 
// Energy balance; from Ex. 1.3, Comment 1 
-q'cv - q'rad + Edot'g = 0 
q'cv = pi*D*h*(T - Tinf) 
q'rad = pi*D*eps*sigma*(T^4 - Tsur^4) 
sigma = 5.67e-8 
 
// The generation term has the form 
Edot'g = I^2*R'e 
qdot = I^2*R'e / (pi*D^2/4) 
 
// Input parameters 
D = 0.001 
Tsur = 300 
T_C = T – 273 // Representing temperature in Celsius units using _C subscript 
eps = 0.8 
Tinf = 300 
h = 100 
//h = 50 // Values of coefficient for parameter study 
//h = 250 
I = 5.2 // For graph, sweep over range from 0 to 10 A 
//I = 4 // For evaluation of heat rates with h = 50 W/m^2.K 
R'e = 0.4 
 
/* Base case results: I = 5.2 A with h = 100 W/m^2.K, find T = 60 C (Comment 2 case). 
Edot'g T T_C q'cv q'rad qdot D I R'e 
 Tinf Tsur eps h sigma 
10.82 332.6 59.55 10.23 0.5886 1.377E7 0.001 5.2 0.4 
 300 300 0.8 100 5.67E-8 */ 
 
/* Results: I = 4 A with h = 50 W/m^2.K, find q'cv = 5.7 W/m and q'rad = 0.67 W/m 
Edot'g T T_C q'cv q'rad qdot D I R'e
 Tinf Tsur eps h sigma 
6.4 336.5 63.47 5.728 0.6721 8.149E6 0.001 4 0.4 
 300 300 0.8 50 5.67E-8 */ 
 
PROBLEM 1.46 
 
KNOWN: Long bus bar of prescribed diameter and ambient air and surroundings temperatures. 
Relations for the electrical resistivity and free convection coefficient as a function of temperature. 
 
FIND: (a) Current carrying capacity of the bus bar if its surface temperature is not to exceed 65°C; 
compare relative importance of convection and radiation exchange heat rates, and (b) Show 
graphically the operating temperature of the bus bar as a function of current for the range 100 ≤ I ≤ 
5000 A for bus-bar diameters of 10, 20 and 40 mm. Plot the ratio of the heat transfer by convection tothe total heat transfer for these conditions. 
 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) Steady-state conditions, (2) Bus bar and conduit are very long in direction 
normal to page, (3) Uniform bus-bar temperature, (4) Radiation exchange between the outer surface of 
the bus bar and the conduit is between a small surface and a large enclosure. 
 
PROPERTIES: Bus-bar material, ( )[ ]e e,o o1 T T ,ρ ρ α= + − e,o 0.0171 m,ρ µ= Ω ⋅ oT 25 C,= ° 
10.00396 K .α −= 
 
ANALYSIS: An energy balance on the bus-bar for a unit length as shown in the schematic above has 
the form 
 in out genE E E 0′ ′ ′− + =� � � 
 
2
rad conv eq q I R 0′ ′ ′− − + = 
 ( ) ( )4 4 2sur e cD T T h D T T I / A 0εpi σ pi ρ∞− − − − + = 
where 2e e c cR / A and A D / 4.ρ pi′ = = Using the relations for ( )e Tρ and ( )h T, D , and substituting 
numerical values with T = 65°C, find 
( ) [ ] [ ]( )4 48 2 4 4radq 0.85 0.020m 5.67 10 W / m K 65 273 30 273 K 223W / mpi −′ = × × ⋅ + − + = < 
 ( )( )2convq 7.83W / m K 0.020m 65 30 K 17.2 W / mpi′ = ⋅ − = < 
where ( ) ( )0.25 0.251.75 1.25 2h 1.21W m K 0.020m 65 30 7.83W / m K−− −= ⋅ ⋅ − = ⋅ 
 ( ) ( )22 2 6 2 5 2eI R I 198.2 10 m / 0.020 m / 4 6.31 10 I W / mpi− −′ = × Ω⋅ = × 
where ( )6 1e 0.0171 10 m 1 0.00396 K 65 25 K 198.2 mρ µ− − = × Ω⋅ + − = Ω⋅   
The maximum allowable current capacity and the ratio of the convection to total heat transfer rate are 
 ( )cv cv rad cv totI 1950 A q / q q q / q 0.072′ ′ ′ ′ ′= + = = < 
For this operating condition, convection heat transfer is only 7.2% of the total heat transfer. 
 
(b) Using these equations in the Workspace of IHT, the bus-bar operating temperature is calculated 
and plotted as a function of the current for the range 100 ≤ I ≤ 5000 A for diameters of 10, 20 and 40 
mm. Also shown below is the corresponding graph of the ratio (expressed in percentage units) of the 
heat transfer by convection to the total heat transfer, cv totq / q .′ ′ 
 Continued ….. 
PROBLEM 1.46 (Cont.) 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
COMMENTS: (1) The trade-off between current-carrying capacity, operating temperature and bar 
diameter is shown in the first graph. If the surface temperature is not to exceed 65°C, the maximum 
current capacities for the 10, 20 and 40-mm diameter bus bars are 960, 1950, and 4000 A, 
respectively. 
 
(2) From the second graph with cv totq / q′ ′ vs. T, note that the convection heat transfer rate is always a 
small fraction of the total heat transfer. That is, radiation is the dominant mode of heat transfer. Note 
also that the convection contribution increases with increasing diameter. 
 
(3) The Workspace for the IHT program to perform the parametric analysis and generate the graphs is 
shown below. It is good practice to provide commentary with the code making your solution logic 
clear, and to summarize the results. 
 
/* Results: base-case conditions, Part (a) 
I R'e cvovertot hbar q'cv q'rad rhoe D Tinf_C Ts_C 
 Tsur_C eps 
1950 6.309E-5 7.171 7.826 17.21 222.8 1.982E-8 0.02 30 65 
 30 0.85 */ 
 
// Energy balance, on a per unit length basis; steady-state conditions 
// Edot'in - Edot'out + Edot'gen = 0 
-q'cv - q'rad + Edot'gen = 0 
q'cv = hbar * P * (Ts - Tinf) 
P = pi * D 
q'rad = eps * sigma * (Ts^4 - Tsur^4) 
sigma = 5.67e-8 
Edot'gen = I^2 * R'e 
R'e = rhoe / Ac 
rhoe = rhoeo * (1 + alpha * (Ts - To) ) 
To = 25 + 273 
Ac = pi * D^2 / 4 
 
// Convection coefficient 
hbar = 1.21 * (D^-0.25) * (Ts - Tinf)^0.25 // Compact convection coeff. correlation 
// Convection vs. total heat rates 
cvovertot = q'cv / (q'cv + q'rad) * 100 
 
// Input parameters 
D = 0.020 
// D = 0.010 // Values of diameter for parameter study 
// D = 0.040 
// I = 1950 // Base case condition unknown 
rhoeo = 0.01711e-6 
alpha = 0.00396 
Tinf_C = 30 
Tinf = Tinf_C + 273 
Ts_C = 65 // Base case condition to determine current 
Ts = Ts_C + 273 
Tsur_C = 30 
Tsur = Tsur_C + 273 
eps = 0.85 
0 1 0 0 0 2 0 0 0 3 0 0 0 4 0 0 0 5 0 0 0
C u rren t, I (A)
2 0
4 0
6 0
8 0
1 0 0
Ba
r 
te
m
pe
ra
tu
re
, T
s 
(C
)
D = 1 0 m m
D = 2 0 m m
D = 4 0 m m
20 40 60 80 100
Bus bar temperature, T (C)
1
3
5
7
9
11
13
R
a
tio
 
q'
cv
 
/ q
'to
t, 
(%
)
D = 10 mm
D = 20 mm
D = 40 mm
PROBLEM 1.47 
 
KNOWN: Elapsed times corresponding to a temperature change from 15 to 14°C for a reference 
sphere and test sphere of unknown composition suddenly immersed in a stirred water-ice mixture. 
Mass and specific heat of reference sphere. 
 
FIND: Specific heat of the test sphere of known mass. 
 
SCHEMATIC: 
 
 
 
ASSUMPTIONS: (1) Spheres are of equal diameter, (2) Spheres experience temperature change 
from 15 to 14°C, (3) Spheres experience same convection heat transfer rate when the time rates of 
surface temperature are observed, (4) At any time, the temperatures of the spheres are uniform, 
(5) Negligible heat loss through the thermocouple wires. 
 
PROPERTIES: Reference-grade sphere material: cr = 447 J/kg K. 
 
ANALYSIS: Apply the conservation of energy requirement at an instant of time, Eq. 1.11a, after 
a sphere has been immersed in the ice-water mixture at T∞. 
 
 in out stE E E− =� � � 
 
 conv
dTq Mc
dt
− = 
 
where ( )conv sq h A T T .∞= − Since the temperatures of the spheres are uniform, the change in 
energy storage term can be represented with the time rate of temperature change, dT/dt. The 
convection heat rates are equal at this instant of time, and hence the change in energy storage 
terms for the reference (r) and test (t) spheres must be equal. 
 
 r r t t
r t
dT dTM c M c
dt dt
 
=   
 
Approximating the instantaneous differential change, dT/dt, by the difference change over a short 
period of time, ∆T/∆t, the specific heat of the test sphere can be calculated. 
 
 
( ) ( )
t
15 14 K 15 14 K
0.515 kg 447 J / kg K 1.263kg c
6.35s 4.59s
− −
× ⋅ = × × 
 
 tc 132 J / kg K= ⋅ < 
 
COMMENTS: Why was it important to perform the experiments with the reference and test 
spheres over the same temperature range (from 15 to 14°C)? Why does the analysis require that 
the spheres have uniform temperatures at all times? 
 
 
PROBLEM 1.48 
 
KNOWN: Inner surface heating and new environmental conditions associated with a spherical shell of 
prescribed dimensions and material. 
 
FIND: (a) Governing equation for variation of wall temperature with time. Initial rate of temperature 
change, (b) Steady-state wall temperature, (c) Effect of convection coefficient on canister temperature. 
 
SCHEMATIC: 
 
 
ASSUMPTIONS: (1) Negligible temperature gradients in wall, (2) Constant properties, (3) Uniform, 
time-independent heat flux at inner surface. 
 
PROPERTIES: Table A.1, Stainless Steel, AISI 302: ρ = 8055 kg/m3, cp = 510 J/kg⋅K. 
 
ANALYSIS: (a) Performing an energy balance on the shell at an instant of time, in out stE E E− =� � � . 
Identifying relevant processes and solving for dT/dt, 
 ( ) ( )( ) ( )2 2 3 3i o o pi i4 dTq 4 r h 4 r T T r r c3 dtpi pi ρ pi∞′′ − − = − 
 ( ) ( )2 2i oi3 3p o i
dT 3 q r hr T T
dt c r rρ
∞
 ′′= − −  
−
. 
Substituting numerical values for the initial condition, find 
 
( ) ( ) ( )
( ) ( )
2 25
2 2
3 3 3i
3
W W3 10 0.5m 500 0.6m 500 300 K
dT m m K
kg Jdt 8055 510 0.6 0.5 m
kg Km
 
− −  ⋅ 
=  
−  ⋅
 
 
i
dT 0.089 K/s
dt

= − . < 
(b) Under steady-state conditions with stE� = 0, it follows that 
 ( ) ( )( )2 2i oiq 4 r h 4 r T Tpi pi ∞′′ = − 
 
 Continued ….. 
 
 
PROBLEM 1.48 (Cont.) 
 
 
2